Let y = y(x) be the solution of the differential equation $$x\tan \left( {{y \over x}} \right)dy = \left( {y\tan \left( {{y \over x}} \right) - x} \right)dx$$, $$ - 1 \le x \le 1$$, $$y\left( {{1 \over 2}} \right) = {\pi \over 6}$$. Then the area of the region bounded by the curves x = 0, $$x = {1 \over {\sqrt 2 }}$$ and y = y(x) in the upper half plane is :

Let $$A = [{a_{ij}}]$$ be a 3 $$\times$$ 3 matrix, where $${a_{ij}} = \left\{ {\matrix{ 1 & , & {if\,i = j} \cr { - x} & , & {if\,\left| {i - j} \right| = 1} \cr {2x + 1} & , & {otherwise.} \cr } } \right.$$

Let a function f : R $$\to$$ R be defined as f(x) = det(A). Then the sum of maximum and minimum values of f on R is equal to:

The number of real roots of the equation $${\tan ^{ - 1}}\sqrt {x(x + 1)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = {\pi \over 4}$$ is :

Let y = y(x) be the solution of the differential equation $${e^x}\sqrt {1 - {y^2}} dx + \left( {{y \over x}} \right)dy = 0$$, y(1) = $$-$$1. Then the value of (y(3))² is equal to :